# 4.4: Inverse Functions

- Page ID
- 528

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An inverse function is a function that undoes another function: If an input \(x\) into the function \(f\) produces an output \(y\), then putting \(y\) into the inverse function \(g\) produces the output \(x\), and vice versa.

Definition: Inverse Functions

Let \(f(x)\) be a 1-1 function then \(g(x)\) is an inverse function of \(f(x)\) if

\[ f(g(x)) = g(f(x)) = x. \]

Example \(\PageIndex{1}\)

For

\[ f(x) = 2x - 1 \nonumber \]

\[ f^{ -1}(x) = \dfrac{1}{2} x +\dfrac{1}{2} \nonumber \]

since

\[ f(f^{ -1}(x) ) = 2[\dfrac{1}{2} x +\dfrac{1}{2}] - 1 = x \nonumber \]

and

\[ f ^{-1}(f(x)) = \dfrac{1}{2} [2x - 1] + \dfrac{1}{2} = x. \nonumber \]

## The Horizontal Line Test and Roll's Theorem

**Note** that if \(f(x)\) is differentiable and the horizontal line test fails then

\[ f(a) = f(b) \]

and *Rolls theorem* implies that there is a \(c\) such that

\[ f '(c) = 0.\]

A partial converse is also true:

Theorem (Roll's Theorem)

If \(f\) is differentiable and \(f '(x)\) is always non-negative (or always non-positive) then \(f(x)\) has an inverse.

Example \(\PageIndex{2}\)

\[ f(x) = x^3 + x - 4 \nonumber \]

has an inverse since

\[f'(x) = 3x^2 + 1 \nonumber \]

which is always positive.

## Continuity and Differentiability of the Inverse Function

Theorem (Continuity and Differentiability)

- \(f\) continuous implies that \(f^{ -1}\) is continuous.
- \(f\) increasing implies that \(f^{ -1}\) is increasing.
- \(f\) decreasing implies that \(f^{ -1}\) is decreasing.
- \(f\) differentiable at \(c\) and \(f '(c) \neq 0\) implies that \(f^{ -1}\) is differentiable at \(f (c)\).
- If \(g(x)\) is the inverse of the differentiable \(f(x)\) then

\[ g'(x) = \dfrac{1}{f '(g(x))}.\]

if \(f '(g(x)) \neq 0\).

Proof

**Proof of (5)**

Since

\[ f (g(x)) = x \nonumber \]

we differentiate implicitly:

\[\dfrac{d}{dx} f(g(x)) = \dfrac{d}{dx} x.\nonumber \]

Using the chain rule

\[ y =f(u), u = g(x)\nonumber \]

\[ \begin{align*} \dfrac{dy}{x} &= \dfrac{dy}{dy} \dfrac{du}{dx} \nonumber \\[4pt] &= f '(u) g'(x) = f '(g(x)) g'(x). \end{align*}\]

So that

\[ f '(g(x)) g'(x) = 1. \nonumber \]

Dividing, we get:

\[g'(x) = \dfrac{1}{f'(g(x))}.\nonumber \]

\(\square\)

Example \(\PageIndex{3}\)

For \(x > 0\), let

\[ f(x) = x^2\nonumber \]

and

\[ g(x) = \sqrt{x}\nonumber \]

be its inverse, then

\[ g'(x) = \dfrac{1}{2\sqrt{x}}.\nonumber \]

Note that

\[ \dfrac{d}{dx} \sqrt{x} = \dfrac{d}{dx} x^{\frac{1}{2}} = \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}.\nonumber \]

Exercises

1. Let

\[ f(x) = x^3 + x - 4.\nonumber \]

Find

\[ \dfrac{d}{dx} f^{-1}(-4). \nonumber \]

2. Let

\[f(x) =\int_2^x \dfrac{1}{1+x^3} dx\nonumber \]

Find

\[ \dfrac{d}{dx} f^{-1}(0). \nonumber \]

## Outside Links

- http://en.wikipedia.org/wiki/Rolle%27s_theorem
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.